UVA

题意：一个人要逃离迷宫，迷宫中有几处起火了。问能否逃出迷宫，能输出最小步数，不能输出"IMPOSSIBLE"。迷宫的边缘都是出口。

思路：BFS，火和人同时进行BFS即可。注意火不止一处。我没开标记数组，而是直接把走过的地方直接标记为‘#’。

#include<bits/stdc++.h>using namespace std;const int INF = 0x3f3f3f;const int MAXN = 1005;int n, m, fire_num, dir[4][2] = {{1,0}, {0, 1}, {-1, 0}, {0,-1}};struct Point{    int x, y, step;    bool op;//people or fire}fire[MAXN], people, NOW, NEXT;char MAP[MAXN][MAXN];int bfs(){    queue<Point> q;    for(int i = 0; i < fire_num; i++) q.push(fire[i]);//火源入队    q.push(people);//人入队    while(!q.empty())    {        NOW = q.front();        q.pop();        if(NOW.op && (NOW.x == 0 || NOW.y == 0 || NOW.x == n - 1 || NOW.y == m - 1)) return NOW.step + 1;        for(int i = 0; i < 4; i++)        {            int X = NOW.x + dir[i][0], Y = NOW.y + dir[i][1];            if(X >= 0 && Y >= 0 && X < n && Y < m && MAP[X][Y] != '#')            {                MAP[X][Y] = '#';                NEXT.x = X; NEXT.y = Y;                NEXT.step = NOW.step + 1; NEXT.op = NOW.op;                q.push(NEXT);            }        }    }    return -1;}int main(){    int T;    scanf("%d", &T);    while (T--)    {        fire_num = 0;        scanf("%d%d", &n, &m);        for (int i = 0; i < n; i++)        {            for(int j = 0; j < m; j++)            {                cin >> MAP[i][j];                if(MAP[i][j] == 'J')                {                    people.x = i; people.y = j;                    people.step = 0; people.op = true;                }                else if(MAP[i][j] == 'F')                {                    fire[fire_num].x = i; fire[fire_num].y = j;                    fire[fire_num].step = 0; fire[fire_num].op = false;                    fire_num++;                }            }        }        int ans = bfs();        if(ans != -1) printf("%d\n",ans);        else printf("IMPOSSIBLE\n");    }    return 0;}/*24 4#####JF##..##..#3 3####J.#.F*/